
Re: Order Isomorphic
Posted:
Jan 31, 2013 2:00 AM


On Jan 30, 11:01 pm, William Elliot <ma...@panix.com> wrote: > Is every infinite subset S of omega_0 with the inherited order, > order isomorphic to omega_0?
Yes, of course. (And you don't need the subscript; omega without a subscript means omega_0, the first infinite ordinal.)
> Yes. S is an ordinal, a denumerable ordinal. > Let eta be the order type of S.
No, S is *not* an ordinal, unless S = omega. S is a subset of an ordinal, so it's a wellordered set, so it's isomorphic to an ordinal, so it's order type is an ordinal.
> [. . .] > Does the same reasoning hold to show that an uncountable subset > of omega_1 with the inherited order is order isomorphic to omega_1.
Yes, of course. The ordinal omega_1 is the unique ordinal such that (a) it';s uncountable, and (b) each of its proper initial segments is countable. Any wellordered set with those two properties is isomorphic to omega_1. If S is an uncountable subset of omega_1, then S (with the natural order) has those two properties.
More generally (and just as trivially), if kappa is an initial ordinal, and if S is a subset of kappa which has the the same cardinality as kappa, then S is orderisomorphic to kappa.

